Question

Suppose an airline claims that its flights are consistently on time. It claims that the average delay is so consistent that the standard deviation is no more than 12.25 minutes. Doubting this claim, a disgruntled traveler calculates the delays for his next 25 flights. The standard deviation for his next 25 flights is 15 minutes. Test at the 5% significance level.

Calculate the test statistic. Round to two decimal places.

What are the degrees of freedom?

What is the p-value? Round to 4 decimal places.

Based on the p-value, do you reject the null hypothesis? Based on the p-value, do you reject the null hypothesis?

Answer 1

i am using minitab to solve the problem.

steps :-

stat basic statistics 1 variance select sample standard deviation from the drop down menutype 25 in sample size and 15 in sample standard deviation tick perform hypothesis test select hypotheized standard deviation from the drop doqn menu in value type 12.25optionsin confidence level type 95select the alternative hypothesis as standard deviation>hypothesizedstandard deviationokok.

to round off any value to needed decimal places :-

right click on the value decimal places select the needed place(here, we will select 4 in p value )

*** SOLUTION***

hypothesis:-

the test statistic be:-

the degrees of freedom = 24

the p value = 0.0551

decision:-

p value = 0.0551 > 0.05 (alpha)

we fail to reject the null hypothesis.

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